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3 years ago

How much energy is required to raise the temperature of 10.7 grams of gaseous helium from 22.1 °C to 39.4 °C ?

Chemistry

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

Q = 2640.96 J

Explanation:

Given data:

Mass of He gas = 10.7 g

Initial temperature = 22.1°C

Final temperature = 39.4°C

Heat absorbed = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 39.4°C - 22.1°C

ΔT = 17.3°C

Q = 10.7 g× 14.267 J/g.°C × 17.3°C

Q = 2640.96 J

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$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

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Putting values in above equation, we get:

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For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

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T = temperature = 500 K

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Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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3 years ago

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